# Matrix e^(tA)

• Mar 6th 2011, 10:55 AM
Borkborkmath
Matrix e^(tA)
I'm trying to figure out if I am doing this correctly.

Given the matrix A:
[1 -.5]
[.5 2]
write down e^(tA) and calculate e^(tA)X, X = (1,1)

So far what I did was, I found the eigenvalues (1.5 repeated) and the eigenvectors ( [-1 1], [1 1]). (The eigenvectors are transposed)
Then I set e^(tA) = T^(-1)AT, where T = [eigenvector1, eigenvector 2]

[1.75 .75]
[0 1.5]

I Feel that I do not have the correct equation for e^(tA), can anyone help me?
• Mar 6th 2011, 11:24 AM
Jester
For a 2 X 2 system, how can you have two eigenvectors for a repeated eigenvalue?
• Mar 6th 2011, 11:43 AM
Borkborkmath
The second eigenvector is the adjoint vector.
I calculated it from (A -1.5I)W = V, where V is the eigenvector of the eigenvalue 1.5
• Mar 6th 2011, 12:56 PM
HallsofIvy
First, it is very easy for a 2 by 2 matrix, having only one eigenvalue, to have two independent eigenvectors- the matrix
$\displaystyle \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}$
has the single eigenvalue 3 and has <1, 0> and <0, 1> as independent eigenvectors.

However, this matrix does not. It has only multiples of <1, -1> as eigenvectors. <1, 1> is a "generalized eigenvector".

But "$\displaystyle T^{-1}AT$", even for a diagonalizable matrix, is NOT "$\displaystyle e^{At}$". Rather, $\displaystyle T^{-1}AT= D$ where D is the diagonal matrix having the eigenvalues on the diagonal. Then $\displaystyle e^{At}= Te^DT^{-1}$ where $\displaystyle e^{D}$ is the diagonal matrix having e to the eigenvalue powers on the diagonal.

That is, if $\displaystyle D= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$ then $\displaystyle e^D= \begin{bmatrix}e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2} \end{bmatrix}$.

However, when we do not have a "complete set of eigenvectors" as here, we cannot diagonalize the matrix. What we can do is write it in "Jordan Normal Form" which, here, would be $\displaystyle J= \begin{bmatrix}\frac{3}{2} & 1 \\ 0 & \frac{3}{2}\end{bmatrix}$.

And if $\displaystyle B= \begin{bmatrix}\lambda_1 & 1 \\ 0 & \lambda_1\end{bmatrix}$ then
$\displaystyle e^{Bt}= \begin{bmatrix}e^{\lambda_1t} & te^{\lambda_1t} \\ 0 & e^{\lambda_1t}\end{bmatrix}$
• Mar 6th 2011, 02:29 PM
Borkborkmath
You mentioned that e^D will not be the diagonal matrix with the eigenvalues as the diagonal elements.

Am I supposed to solve e^(tA) = Te^(D)T^(-1)
where D:
(1.5 1)
(0 1.5)

So that e^D:
(e^(1.5t) te^1.5t)
(0 e^(1.5t))
or
e^D:
(e^(1.5) e^(1.5))
(0 e^(1.5))?

I did it the first way and solved it out to e^(tA)=
(-.5te^(1.5t)+e^(1.5t) -.5te^(1.5t))
(.5te^(1.5t) .5te^(1.5t)+e^(1.5t))
• Mar 7th 2011, 12:20 AM
Opalg
Quote:

Originally Posted by Borkborkmath
I did it the first way and solved it out to $\displaystyle e^{tA}= \begin{bmatrix} -.5te^{1.5t}+e^{1.5t} & -.5te^{1.5t} \\ .5te^{1.5t} & .5te^{1.5t}+e^{1.5t} \end{bmatrix}$

Yes, that's the same answer that I got! (Clapping)